Diodes
A Diode Problem with Unterminated LEads
Goal
Diodes
Solve the for V_{out} assuming the large signal model for the diode
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If V_{in}=1.7 V in the diode circuit below, what is V_{out}, assuming the large signal model for the diode V_{on}=0.7 V
\begin{align} &a.& 0 V \\ &b.& 0.7 V\\ &c.& 1 V\\ &d.& 1.7 V\\ &e.& 2.4 V\\ \end{align}
\begin{align} &a.& 0 V \\ &b.& 0.7 V\\ &c.& 1 V\\ &d.& 1.7 V\\ &e.& 2.4 V\\ \end{align}
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What should we do about those terminals?
Part 1
Guess diode mode
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The first thing to do when a diode is connected to a DC source is guess if it will be on or off.
Since the + end of the source is connected to the back of the diode, and the source is larger than 0.7V, we'll take "on" as our 1st guess
Since the + end of the source is connected to the back of the diode, and the source is larger than 0.7V, we'll take "on" as our 1st guess
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1A
Compute values using KVL
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Since we guessed the diode was ON, replace the diode with a 0.7V drop generic element.
We want V_{out}, but it isn't an element we have. But whatever V_{out} is connected to, we know that it has the same voltage as the resistor. They are in parallel, so V_{out}=V_R
We want V_{out}, but it isn't an element we have. But whatever V_{out} is connected to, we know that it has the same voltage as the resistor. They are in parallel, so V_{out}=V_R
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Now we can use a KVL loop to find the voltage across the resistor
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Why this loop?
I choose a clockwise loop so that V_{in} will be positive in my KVL equation
+V_{in} - V_{on} -V_{out}=0
Apply KVL. The source is positive, the voltage drop across the diode and resistor are both negative.
V_{in}-V{on}=V_{out}
Subtract the desired variable V_{out} over to the other side to isolate
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V_{out}=1.7-0.7=1.0 V
Just plug it in to solve.
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What about what's attached to the terminals?